Problem 22 - Calculus: Early Transcendentals

Question

Continuity at a point Determine whether the following functions are continuous at a. Use the continuity checklist to justify your answer. $$f(x)=\left\\{\begin{array}{ll}\frac{x^{2}-4 x+3}{x-3} & \text { if } x \neq 3 \\\2 & \text { if } x=3\end{array} ; a=3\right.$$

Step-by-Step Solution

Answer

Solution: 1. The function is defined at \(a=3\) as \(f(x)=2\). 2. The limit of the function as \(x\) approaches \(3\) exists and is \(2\). 3. The limit of the function as \(x\) approaches \(3\) equals the function's value at \(3\). Since all three conditions are met, the function is continuous at \(a=3\).

Step 1: Check if the function is defined at \(a=3\).

The function is defined as \(f(x)=2\) when \(x=3\). So, the function is defined at \(a=3\).

Step 2: Check if the limit exists as \(x\) approaches \(a=3\).

To check for the existence of the limit, we need to simplify the other part of the function definition (when \(x \neq 3\)) and then find the limit as \(x\) approaches \(3\). Let's simplify the function: $$\frac{x^{2}-4 x+3}{x-3}$$ Now, factoring the numerator: $$\frac{(x-3)(x-1)}{x-3}$$ We see that there's a common factor in the numerator and the denominator \((x-3)\), which we can now cancel: $$\frac{(x-3)(x-1)}{x-3} = x-1$$ Now, let's find the limit of the simplified function as \(x\) approaches \(3\): $$\lim_{x \to 3} (x-1)$$

Step 3: Calculate the limit as \(x\) approaches \(a=3\).

Since the simplified function is a linear function, the limit exists and is simply the value of the function at \(x=3\): $$\lim_{x \to 3} (x-1) = (3-1)=2$$

Step 4: Check if the limit equals the function's value at \(a=3\).

We've found that the limit of the function as \(x\) approaches \(3\) is equal to \(2\). We also know that the function is defined as \(f(x)=2\) when \(x=3\). Thus, the limit of the function as \(x\) approaches \(3\) equals the function's value at \(3\). Since all three parts of the continuity checklist are satisfied, we can conclude that the function is continuous at \(a=3\).

Key Concepts

Understanding Limits

Limits help us understand how a function behaves as the input approaches a certain point, even if it never actually reaches that point. In the context of our problem, we investigate what happens to the function as \(x\) gets closer to \(3\). We want to know if the outputs of the function are getting close to a specific number as \(x\) approaches \(3\).
This process involves evaluating the function as closely as possible to \(x = 3\) without actually using \(x = 3\). By doing this, we determine whether a "limit" exists at that point. If the function does indeed stabilize to a number (which it does in this exercise), we can say the limit exists. Understanding how limits work is crucial for tackling problems related to continuity.

• The limit tells us the behavior of a function as \(x\) approaches a certain value.
• If the value of the function is consistent as \(x\) approaches the point, the limit exists.

Factoring in Algebra

Factoring simplifies polynomial expressions by breaking them down into smaller 'factorable' terms. In our exercise, we applied factoring to simplify a rational expression in order to find the limit. For the expression \(x^2 - 4x + 3\), we transformed it into a product of simpler terms: \((x-3)(x-1)\).
This step helps us easily cancel out terms that might create undefined conditions, like dividing by zero. Factoring aids in simplifying complex calculations and is a valuable tool in making problems involving polynomials more approachable.

• Factoring breaks down expressions into easier-to-manage parts.
• It lets us simplify computations and solve equations more effectively.

Function Simplification

Simplifying a function enables us to assess it more easily, especially when dealing with its limits and continuity. In this exercise, once the numerator \(x^2 - 4x + 3\) was factored into \((x-3)(x-1)\), we canceled the common term \((x-3)\) with the denominator.
After canceling, the expression reduced to \(x-1\). This simplified form shows that except at any points that were removed, the function behaves normally and makes evaluating its limit straightforward. Simplifying functions is an essential practice that results in cleaner, more understandable expressions.

• Simplifying removes unnecessary complexity from an expression.
• Canceled terms often help clear potential undefined scenarios.

Rational Functions

Rational functions are expressions made by dividing two polynomials. They often present interesting opportunities and challenges when it comes to continuity and limits. In our exercise, the original function \(f(x) = \frac{x^2 - 4x + 3}{x-3}\) is a rational function.
These functions can have points of discontinuity, which occur where the denominator equals zero, as it creates a division by zero problem. To analyze these points, we used factoring and simplification. As shown, once simplified, the function avoids division by zero at \(x = 3\).

• Rational functions consist of a numerator and a denominator, both polynomials.
• These functions can create discontinuities if the denominator values end up being zero.